Theorems For Calculating Limits

In this section, we will learn how to find the limit of the sum, difference, product, and quotient of two functions when their individual limits exist. Additionally, we introduce indeterminate forms, which occur when direct application of these limit laws yields an ambiguous result.

Quick Reference

Limit Law	Formula	Condition
Constant Multiple Law	$lim_{x \to s} [k f (x)] = k lim_{x \to s} f (x)$	Constant $k \in ℝ$
Sum Law	$lim_{x \to s} [f (x) + g (x)] = lim_{x \to s} f (x) + lim_{x \to s} g (x)$	Both limits exist
Product Law	$lim_{x \to s} [f (x) g (x)] = (lim_{x \to s} f (x)) (lim_{x \to s} g (x))$	Both limits exist
Quotient Law	$lim_{x \to s} \frac{f (x)}{g (x)} = \frac{lim_{x \to s} f (x)}{lim_{x \to s} g (x)}$	Provided $lim_{x \to s} g (x) \neq 0$
Basic Indeterminate Forms	$\infty - \infty, 0 \cdot (\pm \infty), \frac{0}{0}, \frac{\pm \infty}{\pm \infty}$	Requires algebraic manipulation or L'Hôpital's Rule

Algebraic Operations on Limits

In this section, for brevity s signifies a, $a^{+}$ , $a^{-}$ , $- \infty$ , or $+ \infty$ where a is a real number.

(Constant Multiple Law) The limit of a constant multiplied by a function is the constant multiplied by the limit of the function:

lim_{x \to s} [k f (x)] = k lim_{x \to s} f (x) .

(Sum Law) The limit of a sum is the sum of the limits: (Because $f (x) - g (x) = f (x) + [- g (x)] = f (x) + (- 1) g (x)$ , it follows from the Sum Law and Constant Multiple Law that $lim_{x \to s} [f (x) - g (x)] = lim_{x \to s} f (x) - lim_{x \to s} g (x)$ .)

lim_{x \to s} [f (x) + g (x)] = lim_{x \to s} f (x) + lim_{x \to s} g (x) .

If you add a function whose limit is a number to a function whose limit is infinity, the limit of the result will be infinity. If one limit is $+ \infty$ and the other one is $- \infty$ , then the limit of the sum can be a number, $+ \infty$ , $- \infty$ , or the sum may oscillate and not approach anything. We say $\infty - \infty$ is an indeterminate form.

(Product Law) The limit of a product is the product of the limits:

lim_{x \to s} [f (x) g (x)] = (lim_{x \to s} f (x)) (lim_{x \to s} g (x)) .

If you multiply a function $f (x)$ whose limit is a number ( $\neq 0$ ) by another function $g (x)$ whose limit is infinity, the limit of the result $f (x) g (x)$ will be infinity, but if that number is negative, the sign of infinity flips. If $lim_{x \to s} f (x) = 0$ and $lim_{x \to s} g (x) = + \infty$ or $- \infty$ , then $lim_{x \to s} [f (x) g (x)]$ depends on the problem. In other words, $0 \cdot (\pm \infty)$ is another indeterminate form.

(Quotient Law) The limit of a quotient is the quotient of the limits provided that the limit of the denominator is not zero:

lim_{x \to s} \frac{f (x)}{g (x)} = \frac{lim_{x \to s} f (x)}{lim_{x \to s} g (x)} (provided lim_{x \to s} g (x) \neq 0)

If $lim_{x \to s} g (x) = 0$ but $lim_{x \to s} f (x) = L \neq 0$ , then

\lim_{x\to s} \frac{f(x)}{g(x)} = \begin{cases} +\infty & \text{if } L \text{ and } g(x) \text{ have the same signs} \\ -\infty & \text{if } L \text{ and } g(x) \text{ have opposite signs} \\ \text{does not exist} & \text{if the sign of } g(x) \text{ changes} \end{cases}.

If the limits of the numerator and denominator are both zero or are both infinity, the result may vary from one problem to another; that is, $\frac{0}{0}$ and $\frac{\pm \infty}{\pm \infty}$ are indeterminate forms.

Proof of the Limit Laws

We only prove the Sum Law, the Product Law, and the Quotient Law as the Constant Multiple Law is a special case of the Product Law when $g (x) \equiv k$ . Also we provide the proof for the cases where x approaches a real number a. However, for the case where x approaches $+ \infty$ or $- \infty$ , proofs can be achieved with minor changes.

Before we start, notice that $lim_{x \to a} f (x) = L$ is equivalent to $lim_{x \to a} [f (x) - L] = 0$ .

Sum Law: Let $ϵ > 0$ be given. We must show that there is a $δ > 0$ such that

| f (x) + g (x) - L - M | < ϵ whenever 0 < | x - a | < δ .

Using the triangle inequality ( $| A + B | \leq | A | + | B |$ ):

| [f (x) - L] + [g (x) - M] | \leq | f (x) - L | + | g (x) - M |

Since $lim_{x \to a} f (x) = L$ , there is a $δ_{1} > 0$ such that $| f (x) - L | < \frac{ϵ}{2}$ whenever $0 < | x - a | < δ_{1}$ . Similarly since $lim_{x \to a} g (x) = M$ , there is a $δ_{2} > 0$ such that $| g (x) - M | < \frac{ϵ}{2}$ whenever $0 < | x - a | < δ_{2}$ .

If $δ = min {δ_{1}, δ_{2}}$ then we have $| f (x) - L | < \frac{ϵ}{2}$ and $| g (x) - M | < \frac{ϵ}{2}$ and hence

| f (x) + g (x) - L - M | \leq | f (x) - L | + | g (x) - M | < \frac{ϵ}{2} + \frac{ϵ}{2} = ϵ

whenever $0 < | x - a | < δ$ .

Product Law: First we prove a special case where $M = 0$ . In other words, we prove that if $f (x) \to L$ and $g (x) \to 0$ as $x \to a$ , then

lim_{x \to a} [f (x) g (x)] = 0.

To this end, we need to show that if $ϵ > 0$ is given, there is a $δ > 0$ such that $| f (x) g (x) | < ϵ$ whenever $0 < | x - a | < δ$ . Since $lim_{x \to a} f (x) = L$ , there must be a $δ_{1} > 0$ such that $| f (x) - L | < 1$ whenever $0 < | x - a | < δ_{1}$ (here we have taken $ϵ = 1$ ).

Therefore, if $0 < | x - a | < δ_{1}$ , then $| f (x) | = | f (x) - L + L | \leq | f (x) - L | + | L | < 1 + | L |$ , and

|f(x) g(x)| = |f(x)| |g(x)| < (1 + |L|) |g(x)|. \tag{*}

Since $lim_{x \to a} g (x) = 0$ , there must be a $δ_{2} > 0$ such that

|g(x)| < \frac{\epsilon}{1 + |L|} \quad \text{whenever} \quad 0 < |x - a| < \delta_2. \tag{**}

If $δ = min {δ_{1}, δ_{2}}$ , then both (*) and (**) are valid whenever $0 < | x - a | < δ$ and hence for such x: $| f (x) g (x) | < ϵ$ .

Now consider the case where $M \neq 0$ . In this case, we simply need to show $lim_{x \to a} [f (x) g (x) - L M] = 0$ . Since $f (x) - L \to 0$ and $g (x) - M \to 0$ as $x \to a$ , and

f (x) g (x) - L M = [f (x) - L] g (x) + [g (x) - M] L

it follows from the case that we just proved that $f (x) g (x) - L M \to 0$ as $x \to a$ . This completes the proof of the Product Law.

Quotient Law: For the Quotient Law, if we prove that $lim_{x \to a} \frac{M}{g (x)} = 1$ , then since $\frac{f (x)}{g (x)} = \frac{f (x)}{M} \cdot \frac{M}{g (x)}$ , it follows from the Product Law that:

lim_{x \to a} (\frac{f (x)}{M} \frac{M}{g (x)}) = lim_{x \to a} \frac{f (x)}{M} \cdot lim_{x \to a} \frac{M}{g (x)} = \frac{1}{M} lim_{x \to a} f (x) \times 1 = \frac{L}{M} .

To prove that $lim_{x \to a} \frac{M}{g (x)} = 1$ , let $h (x) = \frac{g (x)}{M}$ . Now we must show that for a given $ϵ > 0$ , there is a $δ > 0$ such that $| \frac{1}{h (x)} - 1 | < ϵ$ whenever $0 < | x - a | < δ$ . But $| \frac{1}{h (x)} - 1 | = | \frac{1 - h (x)}{h (x)} | = \frac{| h (x) - 1 |}{| h (x) |}$ . Since $lim_{x \to a} h (x) = 1$ , there is a $δ > 0$ such that both $| h (x) - 1 | < \frac{ϵ}{2}$ and $| h (x) - 1 | < \frac{1}{2}$ whenever $0 < | x - a | < δ$ .

Now notice that $| h (x) - 1 | < \frac{1}{2} ⟺ - \frac{1}{2} < h (x) - 1 < \frac{1}{2} ⟺ \frac{1}{2} < h (x) < \frac{3}{2}$ . And $\frac{1}{2} < h (x)$ implies $\frac{1}{| h (x) |} = \frac{1}{h (x)} < 2$ . Therefore, if $0 < | x - a | < δ$ , we have both $| h (x) - 1 | < \frac{ϵ}{2}$ and $\frac{1}{| h (x) |} < 2$ . This means that $| \frac{1}{h (x)} - 1 | = \frac{| h (x) - 1 |}{| h (x) |} < 2 \cdot \frac{ϵ}{2} = ϵ$ whenever $0 < | x - a | < δ$ .

Corollaries: Power and Polynomial Limits

Using the Limit Laws established above, together with the linear limit theorem proved in The Precise Definitions of Limits ( $lim_{x \to a} (m x + b) = m a + b$ ), we can deduce two important corollaries that greatly simplify limit calculations.

(Power Law for Positive Integers):
For any positive integer n and any real number a:

lim_{x \to a} x^{n} = a^{n} .

Proof

Recall from The Precise Definitions of Limits that for any linear function $f (x) = m x + b$ , we have $lim_{x \to a} (m x + b) = m a + b$ . Setting $m = 1$ and $b = 0$ , we get the base limit:

lim_{x \to a} x = a .

We now proceed by mathematical induction on n.

Base Case ( $n = 1$ ): For

n = 1

lim_{x \to a} x^{1} = lim_{x \to a} x = a = a^{1}

, which holds by the linear limit theorem.

Induction Step: Assume that the statement holds for

n = k

; that is, assume

lim_{x \to a} x^{k} = a^{k}

. We must show that

lim_{x \to a} x^{k + 1} = a^{k + 1}

.

Using the identity

x^{k + 1} = x^{k} \cdot x

and applying the Product Law (

lim_{x \to a} [f (x) g (x)] = (lim_{x \to a} f (x)) (lim_{x \to a} g (x))

): \begin{aligned} \lim_{x\to a} x^{k+1} &= \lim_{x\to a} \left(x^k \cdot x\right) \\ &= \left(\lim_{x\to a} x^k\right) \left(\lim_{x\to a} x\right) \\ &= a^k \cdot a \qquad (\text{by the induction hypothesis and the base case}) \\ &= a^{k+1}. \end{aligned}

By the principle of mathematical induction, $lim_{x \to a} x^{n} = a^{n}$ for all positive integers $n \in ℕ^{+}$ .

(Direct Substitution for Polynomials)
Let $P (x) = c_{n} x^{n} + c_{n - 1} x^{n - 1} + \dots + c_{1} x + c_{0}$ be any polynomial of degree n with real coefficients $c_{n}, c_{n - 1}, \dots, c_{0}$ . Then for any real number a:

lim_{x \to a} P (x) = P (a) = c_{n} a^{n} + c_{n - 1} a^{n - 1} + \dots + c_{1} a + c_{0} .

Proof

Let $P (x) = c_{n} x^{n} + c_{n - 1} x^{n - 1} + \dots + c_{1} x + c_{0}$ . By repeated application of the Sum Law, the limit of a sum of terms equals the sum of their individual limits:

\begin{aligned} \lim_{x\to a} P(x) &= \lim_{x\to a} \left( c_n x^n + c_{n-1} x^{n-1} + \dots + c_1 x + c_0 \right) \\ &=\lim_{x\to a}(c_n x^n) +\lim_{x\to a}(c_{n-1}x^{n-1}) +\cdots + \lim_{x\to a} c_n \end{aligned}
For each term

c_{k} x^{k}

with

k \geq 1

, we apply the Constant Multiple Law (

lim_{x \to a} [c_{k} f (x)] = c_{k} lim_{x \to a} f (x)

) and Corollary 1 (

lim_{x \to a} x^{k} = a^{k}

lim_{x \to a} (c_{k} x^{k}) = c_{k} lim_{x \to a} x^{k} = c_{k} a^{k} .

For the constant term

c_{0}

, applying the linear limit theorem with

m = 0

and

b = c_{0}

gives

lim_{x \to a} c_{0} = c_{0}

.

Combining all terms yields:

lim_{x \to a} P (x) = c_{n} a^{n} + c_{n - 1} a^{n - 1} + \dots + c_{1} a + c_{0} = P (a) .

Thus, the limit of any polynomial $P (x)$ as $x \to a$ is evaluated by simple direct substitution $P (a)$ .

The Indeterminate Forms

Although the preceding theorems tell us a great deal about the behavior of combined functions, there are four cases about which they are silent. These cases are abbreviated as

\infty - \infty, 0 \cdot (\pm \infty), \frac{0}{0}, \frac{\pm \infty}{\pm \infty} .

Limits of these types are called indeterminate forms. The word indeterminate means that the form alone does not determine the answer. Two limits may have the same form and still behave completely differently: one may converge to a real number, another may tend to $+ \infty$ or to $- \infty$ , and a third may fail to exist at all. Deciding which of these occurs requires further work in each individual case.

Three additional indeterminate forms, $0^{0}$ , $1^{\infty}$ , $\infty^{0}$ , arise from exponentiation.

In arithmetic, $0 / 0$ is undefined and $\infty$ is not a number. The symbols $0 / 0$ , $0 \cdot (\pm \infty)$ , and the others above are therefore not operations to be carried out. Each one is shorthand for a limit in which the two parts behave in the indicated way, and nothing more.

Extended Real Numbers

In advanced mathematics, the extended real number system, denoted by $\overset{―}{ℝ}$ or by $[- \infty, + \infty]$ , is formed by adjoining two elements, $+ \infty$ and $- \infty$ , to the set of real numbers $ℝ$ :

\overset{―}{ℝ} = ℝ \cup {- \infty, + \infty} .

To distinguish the real numbers from the two new symbols, the former are called finite and the latter infinite.

Operation of Negation: $- (+ \infty) := - \infty$ and $- (- \infty) := + \infty$ .

Order Structure: $- \infty < x < + \infty$ for every $x \in ℝ$ .

Arithmetic Operations:

Addition: $x + (+ \infty) = + \infty$ , $x + (- \infty) = - \infty$ , $(+ \infty) + (+ \infty) = + \infty$ , $(- \infty) + (- \infty) = - \infty$ .
Subtraction: $x - (+ \infty) = - \infty$ , $x - (- \infty) = + \infty$ .
Multiplication ( $x > 0$ ): $x \cdot (+ \infty) = + \infty$ , $x \cdot (- \infty) = - \infty$ .
Multiplication ( $x < 0$ ): $x \cdot (+ \infty) = - \infty$ , $x \cdot (- \infty) = + \infty$ .
Infinite factors: $(+ \infty) (+ \infty) = (- \infty) (- \infty) = + \infty$ , $(+ \infty) (- \infty) = - \infty$ .
Division: $\frac{x}{+ \infty} = \frac{x}{- \infty} = 0$ .

Expressions such as $(+ \infty) + (- \infty)$ , $\frac{+ \infty}{+ \infty}$ , and $0 \cdot (+ \infty)$ are left undefined in $\overset{―}{ℝ}$ .

Why

0^{0}

1^{\infty}

, and

\infty^{0}

are not separate cases

Each of these three forms reduces to $0 \cdot (\pm \infty)$ once we take a logarithm. Suppose $y = h (x) = f (x)^{g (x)}$ where $f (x) > 0$ . Taking the logarithm gives $\ln y = g (x) \ln f (x)$ , so $y = \exp (g (x) \ln f (x))$ .

Form	$f (x) \to$	$\ln f (x) \to$	$g (x) \ln f (x)$ has the form
$0^{0}$	$0^{+}$	$- \infty$	$0 \cdot (- \infty)$
$1^{\infty}$	$1$	$0$	$\infty \cdot 0$
$\infty^{0}$	$+ \infty$	$+ \infty$	$0 \cdot (+ \infty)$

In all three rows the exponent $g (x) \ln f (x)$ is an indeterminate product, so nothing about $y$ can be concluded from the form alone. If that product does tend to a limit $L$ , then continuity of the exponential function gives $y \to e^{L}$ .

Exercises

The graphs of f and g are shown in the following figure. Use the limit laws and evaluate the following limits, if they exist.

$lim_{x \to - 3} [3 f (x) - 2 g (x)]$
$lim_{x \to - 3} \frac{f (x)}{4 g (x)}$
$lim_{x \to 0} [f (x) g (x)]$

Graphs of functions f and g for Exercise 1 — Graphs of $f (x)$ and $g (x)$ .

Answer

(a) $13$ (b) $- \frac{3}{8}$ (c) Does not exist

Solution

(a) By the Sum Law we have $lim_{x \to - 3} [3 f (x) - 2 g (x)] = lim_{x \to - 3} [3 f (x)] + lim_{x \to - 3} [- 2 g (x)]$ . By the Constant Multiple Law: $3 lim_{x \to - 3} f (x) - 2 lim_{x \to - 3} g (x)$ . From the graphs we find $lim_{x \to - 3} f (x) = 3$ and $lim_{x \to - 3} g (x) = - 2$ . Therefore: $3 (3) - 2 (- 2) = 13$ .

(b) By the Constant Multiple Law: $lim_{x \to - 3} \frac{f (x)}{4 g (x)} = \frac{1}{4} lim_{x \to - 3} \frac{f (x)}{g (x)}$ , and by the Quotient Law: $\frac{1}{4} \frac{lim_{x \to - 3} f (x)}{lim_{x \to - 3} g (x)} = \frac{1}{4} \frac{3}{- 2} = - \frac{3}{8}$ .

(c) From the graphs, $lim_{x \to 0} f (x) = - 2$ but $lim_{x \to 0} g (x)$ does not exist because left and right limits differ: $lim_{x \to 0^{-}} g (x) = 2$ and $lim_{x \to 0^{+}} g (x) = 3$ . Left-hand limit: $(- 2) (2) = - 4$ ; right-hand limit: $(- 2) (3) = - 6$ . Because left and right limits are not equal, $lim_{x \to 0} [f (x) g (x)]$ does not exist.

The graphs of f and g are shown in the following figure. Use the limit laws and evaluate the following limits, if they exist.

$lim_{x \to 1} [f (x) - g (x)]$
$lim_{x \to 1} [f (x) g (x)]$
$lim_{x \to 1} \frac{g (x)}{f (x)}$

Graphs of functions f and g for Exercise 2 — Graphs of $f (x)$ and $g (x)$ .

Answer

(a) Does not exist (b) $0$ (c) $0$

Solution

As $lim_{x \to 1} g (x)$ does not exist, following the previous example, we may be tempted to say that none of the above limits exist. However, we will see that only the first one does not exist.

(a) $lim_{x \to 1^{-}} [f (x) - g (x)] = 2 - 0 = 2$ . $lim_{x \to 1^{+}} [f (x) - g (x)] = - 3 - 0 = - 3$ . Because $lim_{x \to 1^{-}} [f (x) - g (x)] \neq lim_{x \to 1^{+}} [f (x) - g (x)]$ , we conclude that $lim_{x \to 1} [f (x) - g (x)]$ does not exist.

(b) $lim_{x \to 1^{-}} [f (x) g (x)] = (2) (0) = 0$ . $lim_{x \to 1^{+}} [f (x) g (x)] = (- 3) (0) = 0$ . Because left and right limits are both equal to 0, $lim_{x \to 1} [f (x) g (x)] = 0$ .

(c) $lim_{x \to 1^{-}} \frac{g (x)}{f (x)} = \frac{0}{2} = 0$ , $lim_{x \to 1^{+}} \frac{g (x)}{f (x)} = \frac{0}{- 3} = 0$ . Because left and right limits exist and are equal, $lim_{x \to 1} \frac{g (x)}{f (x)} = 0$ .

Evaluate the following limits:

$lim_{x \to 1} [(2 x - 1) x^{3}]$
$lim_{x \to - 2} [3 x (2 x - 1)]$
$lim_{x \to 2} [3 (1 - x) (2 x^{2} + 1)]$
$lim_{x \to - 4} (x + 3)^{71}$

Answer

(a) $1$ (b) $30$ (c) $- 27$ (d) $- 1$

Solution

(a) By direct substitution and applying the Limit Laws:

lim_{x \to 1} [(2 x - 1) x^{3}] = [2 (1) - 1] (1)^{3} = (1) (1) = 1.

(b) By direct substitution:

lim_{x \to - 2} [3 x (2 x - 1)] = 3 (- 2) [2 (- 2) - 1] = (- 6) (- 5) = 30.

(c) By direct substitution:

lim_{x \to 2} [3 (1 - x) (2 x^{2} + 1)] = 3 (1 - 2) [2 (2^{2}) + 1] = 3 (- 1) (9) = - 27.

(d) By direct substitution:

lim_{x \to - 4} (x + 3)^{71} = (- 4 + 3)^{71} = (- 1)^{71} = - 1.

Suppose $lim_{x \to a} f (x) = - 1$ and $lim_{x \to a} g (x) = 2$ . Evaluate the following:

What is $lim_{x \to a} [f (x) + g (x)]$ ?
What is $lim_{x \to a} [f (x) g (x)]$ ?
What is $lim_{x \to a} [4 f (x) g (x)]$ ?
What is $lim_{x \to a} [f (x) - 2 g (x)]$ ?
What is $lim_{x \to a} \frac{f (x)}{- g (x)}$ ?
What is $lim_{x \to a} \frac{2 f (x) + 3 g (x)}{f (x) g (x) + (g (x))^{2} + 1}$ ?

Answer

(a) $1$ (b) $- 2$ (c) $- 8$ (d) $- 5$ (e) $\frac{1}{2}$ (f) $\frac{4}{3}$

Solution

Using the limit laws:

(a) By the Sum Law:

lim_{x \to a} [f (x) + g (x)] = lim_{x \to a} f (x) + lim_{x \to a} g (x) = - 1 + 2 = 1.

(b) By the Product Law:

lim_{x \to a} [f (x) g (x)] = (lim_{x \to a} f (x)) (lim_{x \to a} g (x)) = (- 1) (2) = - 2.

(c) By the Constant Multiple Law and Product Law:

lim_{x \to a} [4 f (x) g (x)] = 4 (lim_{x \to a} f (x)) (lim_{x \to a} g (x)) = 4 (- 1) (2) = - 8.

(d) By the Sum/Difference Law and Constant Multiple Law:

lim_{x \to a} [f (x) - 2 g (x)] = lim_{x \to a} f (x) - 2 lim_{x \to a} g (x) = - 1 - 2 (2) = - 5.

(e) By the Quotient Law and Constant Multiple Law:

lim_{x \to a} \frac{f (x)}{- g (x)} = \frac{lim_{x \to a} f (x)}{- lim_{x \to a} g (x)} = \frac{- 1}{- 2} = \frac{1}{2} .

(f) By combining the Sum, Product, and Quotient Laws:

lim_{x \to a} \frac{2 f (x) + 3 g (x)}{f (x) g (x) + (g (x))^{2} + 1} = \frac{2 (- 1) + 3 (2)}{(- 1) (2) + (2)^{2} + 1} = \frac{- 2 + 6}{- 2 + 4 + 1} = \frac{4}{3} .

Can you provide an example of functions $f$ and $g$ such that

lim_{x \to a} \frac{f (x)}{g (x)}

exists, but at least one of the limits $lim_{x \to a} f (x)$ or $lim_{x \to a} g (x)$ fails to exist?

Answer

Yes. For example, let $f (x) = \frac{1}{x}$ and $g (x) = \frac{1}{x}$ near $a = 0$ . Neither limit exists, but $lim_{x \to 0} \frac{f (x)}{g (x)} = 1$ .

Solution

Yes, such functions exist. The Quotient Law states that the limit of a quotient equals the quotient of the limits provided both individual limits exist and the denominator limit is non-zero. If the individual limits fail to exist, the limit of the quotient may still exist.

Example: Let $a = 0$ , and define $f (x) = \frac{1}{x}$ and $g (x) = \frac{1}{x}$ for $x \neq 0$ .

x \to 0

, neither

lim_{x \to 0} f (x)

nor

lim_{x \to 0} g (x)

exists (they approach

\pm \infty

).
However, for all

x \neq 0

\frac{f (x)}{g (x)} = \frac{1 / x}{1 / x} = 1.

Therefore:

lim_{x \to 0} \frac{f (x)}{g (x)} = lim_{x \to 0} 1 = 1,

which exists.

Is it possible that

lim_{x \to a} [f (x) + g (x)]

exists, but both $lim_{x \to a} f (x)$ and $lim_{x \to a} g (x)$ fail to exist?

Answer

Yes. For example, let $f (x) = \frac{1}{x}$ and $g (x) = - \frac{1}{x}$ near $a = 0$ .

Solution

Yes, it is possible. The Sum Law guarantees that if both $lim_{x \to a} f (x)$ and $lim_{x \to a} g (x)$ exist, then $lim_{x \to a} [f (x) + g (x)]$ exists and equals their sum. However, the converse is not true: the sum of two functions whose individual limits fail to exist can still have a limit.

Example 1: Let $a = 0$ , $f (x) = \frac{1}{x}$ , and $g (x) = - \frac{1}{x}$ for $x \neq 0$ .

Neither $lim_{x \to 0} f (x)$ nor $lim_{x \to 0} g (x)$ exists.
However, for all $x \neq 0$ :

f (x) + g (x) = \frac{1}{x} + (- \frac{1}{x}) = 0.

Therefore:

lim_{x \to 0} [f (x) + g (x)] = lim_{x \to 0} 0 = 0,

which exists.

Example 2 (Piecewise Step Functions): Define f(x) = \text{sgn}(x) = \begin{cases} 1 & \text{if } x > 0 \\ -1 & \text{if } x < 0 \end{cases} and $g (x) = - sgn (x)$ .
Neither limit exists as $x \to 0$ because their left and right limits differ ( $1$ vs $- 1$ ). However, $f (x) + g (x) = 0$ for all $x \neq 0$ , so $lim_{x \to 0} [f (x) + g (x)] = 0$ .

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Algebraic Operations on Limits

Corollaries: Power and Polynomial Limits

The Indeterminate Forms

Exercises

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